The acceleration due to gravity at the pols and the equator is `g_(p) and g_(e)` respectively. If the earth is a sphere of radius `R_(E)` and rotating about its axis with angular speed `omega and g_(p)-g_(e)` given by

Options

A. `(omega^(2))/(R )`
B. `R omega^(2)`
C. `R^(2)omega^(2)`
D. `(omega^(2))/(R^(2))`

Answer

B

Explanation

`g'=g[1-(R omega^(2))/(g)cos^(2)lambda]`
at the equator, `lambda = 0`
`therefore g_(E )= g[1-(R omega^(2))/(g)]=g-R omega^(2)`
at the poles, `lambda = 90^(@) therefore g_(P)= g`
`therefore g_(P)-g_(E )=g-g-(R omega^(2))=R omega^(2)`